Allegiant Air G44982 Flight Status : Live Tracking & Updates

Domestic flight Allegiant Air G44982 takes off from Saint Petersburg (PIE) United States to Knoxville (TYS) United States. It's operated by Allegiant Air. The plane leaves St. Petersburg-Clearwater International Airport at 06:30 America/New_York. The flight is expected to land at McGhee Tyson Airport at 08:05 America/New_York. The flight will last about 1 hours 31 minutes.

G44982 - AAY4982 Allegiant Air (G4)
PIE Saint Petersburg United States
Flight PIE to TYS
TYS Knoxville United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on
Departure St. Petersburg-Clearwater International Airport
IATA: PIEICAO: KPIE
Scheduled 06:30
None 3
Arrival McGhee Tyson Airport
IATA: TYSICAO: KTYS
Scheduled 08:05
None 4
G44982 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 31 minutes
  • Flight Distance: 884 kms / 549 miles
Airline
  • Allegiant Air
  • IATA: G4
  • ICAO: AAY
  • Operating Days: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: Normal passenger
  • Seats: 166
  • Freight capacity: 5.7 tons
  • Passenger classes: Economy
More Detail
  • Aircraft: Airbus A320
  • Callsign: AAY982
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Saint Petersburg: Monday 2026-05-25 12:18 PM
  • Current Time in Knoxville: Monday 2026-05-25 12:18 PM

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Allegiant Air Flight G44982 FAQs

What is the scheduled flight duration for Allegiant Air G44982 flight?

On average, nonstop flight takes 1 hour(s) 31 minutes, with the flight distance of 884 km (549 miles).

What type of aircraft is used for the Allegiant Air G44982 flight?

All Allegiant Air G44982 flights are operated using Airbus A320 aircraft.